minimun-path-sum

Problem

Minimum Path Sum

Problem Description

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

Example:

Input:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

Solution

This is a simple DP problem, current sum can either come from up or left, since it can only move down or right.

For each sum, grid[r][c] = min(grid[r-1][c], grid[r][c-1]) + grid[r][c]; then you find the min path to current position (r,c).

after scan the whole grid, return bottom right value.

NOTE: Below implementation, modified original input grid, from algorithm point of view to reduce space complexity to O(1). But from real project, modified original input not recommended, use extra space.

For example:

Minimum Path Sum

Complexity Analysis

Time Complexity: O(N*M)

Space Complexity: O(1)

• N - number of grid rows

• M - number of grid columns

Code

class Solution {
    public int minPathSum(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) return 0;
        int row = grid.length;
        int col = grid[0].length;
        // init first row, since only move down or right, first line can only move from left to right
        for (int c = 1; c < col; c++) {
            grid[0][c] += grid[0][c - 1];
        }
        // init first col, only move down
        for (int r = 1; r < row; r++) {
            grid[r][0] += grid[r - 1][0];
        }
        for (int r = 1; r < row; r++) {
            for (int c = 1; c < col; c++) {
                grid[r][c] += Math.min(grid[r - 1][c], grid[r][c - 1]);
            }
        }
        return grid[row - 1][col - 1];
    }
}